Chapter 7 Triangles | Exercise 7.3 | Class 9 Maths NCERT Question Detailed Explanation

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✅ In this video,
✔️ Class: 9th
✔️ Subject: Maths
✔️ Chapter: 7 Triangles
✔️ Topic Name: Exercise 7.3

✔️ कक्षा: 9वीं
✔️ विषय: गणित
✔️ अध्याय: 7 त्रिकोण
✔️ विषय का नाम: अभ्यास 7.3
✔ kaksha: 9veen
✔ vishay: ganit
✔ adhyaay: 7 trikon
✔ vishay ka naam: abhyaas 7.3

Question Covered :-
1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that(i) ΔABD ≅ ΔACD(ii) ΔABP ≅ ΔACP(iii) AP bisects ∠A as well as ∠D.(iv) AP is the perpendicular bisector of BC.
2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that(i) AD bisects BC                      (ii) AD bisects ∠A.
3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:(i) ΔABM ≅ ΔPQN(ii) ΔABC ≅ ΔPQR
4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

OR

1. Triangle ABC and triangle DBC are two isosceles triangles on the same base BC, and vertices A and D lie on the same side of BC (see Figure 7.39). If line AD is extended to meet BC at point P, show that:
(i) Triangle ABD is congruent to triangle ACD.
(ii) Triangle ABP is congruent to triangle ACP.
(iii) Line segment AP bisects angle A as well as angle D.
(iv) Line segment AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which side AB is equal to side AC. Show that:
(i) Line segment AD bisects side BC.
(ii) Line segment AD bisects angle A.

3. Two sides AB and BC, and the median AM of triangle ABC, are respectively equal to sides PQ and QR, and the median PN of triangle PQR (see Figure 7.40). Show that:
(i) Triangle ABM is congruent to triangle PQN.
(ii) Triangle ABC is congruent to triangle PQR.

4. BE and CF are two equal altitudes of triangle ABC. Using the RHS (Right angle–Hypotenuse–Side) congruence rule, prove that triangle ABC is isosceles.

5. Triangle ABC is an isosceles triangle with AB equal to AC. Draw AP perpendicular to BC to show that angle B is equal to angle C.


1. tribhuj abch aur tribhuj dbch ek hee aadhaar bch par sthit do samadvibaahu tribhuj hain, tatha sheersh a aur d bch ke ek hee taraph sthit hain (chitr 7.39 dekhen). yadi rekha ad ko bch se bindu p par milane ke lie badhaaya jaata hai, to darshaie ki:
(i) tribhuj abd tribhuj achd ke samatuly hai.
(ii) tribhuj abp tribhuj achp ke samatuly hai.
(iii) rekhaakhand ap kon a ke saath-saath kon d ko bhee samadvibhaajit karata hai.
(iv) rekhaakhand ap, bch ka lambavat samadvibhaajak hai.

2. ad ek samadvibaahu tribhuj abch kee oonchaee hai jisamen bhuja ab bhuja ach ke baraabar hai. darshaie ki:
(i) rekhaakhand ad bhuja bch ko samadvibhaajit karata hai.
(ii) rekhaakhand ad kon a ko samadvibhaajit karata hai.

3. tribhuj abch kee do bhujaen ab aur bch, tatha maadhyika am, kramashah tribhuj pq aur qr, tatha tribhuj pqr kee maadhyika pn ke baraabar hain (chitr 7.40 dekhen). darshaie ki:
(i) tribhuj abm tribhuj pqn ke samatuly hai.

(ii) tribhuj abch tribhuj pqr ke samatuly hai.

4. bai aur chf tribhuj abch kee do baraabar oonchaeeyaan hain. rhs (samakon-karn-bhuja) sarvaangasamata niyam ka upayog karake siddh keejie ki tribhuj abch samadvibaahu hai.

5. tribhuj abch ek samadvibaahu tribhuj hai jisamen ab baraabar ach hai. bch par lamb ap kheenchakar darshaie ki kon b, kon ch ke baraabar hai.

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